﻿ Engineering Entrance Exam Question and Answers | Laws of Motion - Zigya

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# Laws of Motion

#### Multiple Choice Questions

21.

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

• more than 3 but less than 6

• more than 6 but less than 9

• more than 9

• more than 9

A.

more than 3 but less than 6

To reverse the direction 869 Views

22.

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

• g

• 2/3g

• g/3

• g/3

B.

2/3g mg - T = ma 514 Views

23.

A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as

• 17.6 kg ms−1

• 17.565 kg ms−1

• 17.56 kg ms−1

• 17.56 kg ms−1

A.

17.6 kg ms−1

P = mv = 3.513 × 5.00 ≈ 17.6

184 Views

24.

For a particle in uniform circular motion, the acceleration an at point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)

• • • • C.   675 Views

25.

Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B? • 4.9 ms-2 in horizontal direction

• 9.8 ms-2 in vertical direction

• zero

• zero

D.

zero

For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of  A with respect to B is
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)

453 Views

26.

A liquid in a beaker has temperature θ(t) at time t and θ0 is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge (θ – θ0) and t is

• • • • A. According to Newton's law of cooling. Hence the plot of ln(θ – θ0) vs t should be a straight line with negative slope.

473 Views

27.

A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

• 0.16 J

• 1.00 J

• 0.67 J

• 0.67 J

C.

0.67 J

m1u1 + m2u2 = (m1 + m2)v
v = 2/3 m/s 183 Views

28.

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

• The statement I is True, Statement II is False.

• The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

• The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

• The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

264 Views

29.

A particle is projected at 60° to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

• K

• 0

• K/2

• K/2

D.

K/2

132 Views

30.

A thin liquid film formed between a U-shaped wire and a light slider supports a weight of 1.5 x10–2N (see figure). The length of the slider is 30 cm and its weight negligible. The surface tension of the liquid film is • 0.0125 Nm-1

• 0.1 Nm-1

• 0.05 Nm-1

• 0.05 Nm-1

D.

0.05 Nm-1 The force of surface tension acting on the slider balances the force due to the weight.

⇒F = 2Tl = w
⇒2T(0.3) = 1.5 x 10–2
⇒T = 2.5 x 10–2 N/m

369 Views